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kvasek [131]
3 years ago
12

X + 5 = 10 A) x = 2 B) x = 3 C) x = 4 D) x = 5

Mathematics
2 answers:
MrRa [10]3 years ago
8 0

D.) x = 5, because when you plug five in for x, 5 + 5 = 10 is correct.

Blababa [14]3 years ago
3 0

Hey there!

            \bold{x+5=10}

  1. \bold{A.)x=2}
  2. \bold{B.)x=3}
  3. \bold{C.)x=4}
  4. \bold{D.)x=5}

<h2>Okay, so first we have to do the <u><em>OPPOSITE</em></u> of \bold{addition} (which is \bold{subtraction})</h2><h3>So, \bold{subtract} by \bold{5} on each of the sides </h3>
  • \bold{x+5-5}  \bold{=10-5}
  • \bold{Cancel:5-5} because that gives you the result of \bold{0} which is <u><em>NOT</em></u> our answer
  • \bold{Keep:10-5} because it gives us the result your "\bold{x-term} "
  • \boxed{\boxed{\bold{Answer:x=5}}}

Good luck on your assignment and enjoy your day!


~\bold{LoveYourselfFirst:)}


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-15x + 45 = 2x - 126 - 33
-15x - 2x = -204
17x = 204
x = 204/17
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In short, Your Answer would be 12

Hope this helps!
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Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

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astra-53 [7]
The answer is 14.4 cm.
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What are the zeros of the polynomial function
myrzilka [38]

Answer:

(b) 0,-5,4

Step-by-step explanation:

f(x)=x^3+x^2-20x\\x^3+x^2-20x=0\\x(x^2+x-20)=0\\x(x+5)(x-4)=0\\x=0, x+5=0, x-4=0\\x=0,x=-5,x=4

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4 years ago
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